[[!toc]] #Q: How do you know that every term in the untyped lambda calculus has a fixed point?# A: That's easy: let `T` be an arbitrary term in the lambda calculus. If `T` has a fixed point, then there exists some `X` such that `X <~~> TX` (that's what it means to *have* a fixed point).
``````let L = \x. T (x x) in
let X = L L in
X ≡ L L ≡ (\x. T (x x)) L ~~> T (L L) ≡ T X``````
Please slow down and make sure that you understand what justified each of the equalities in the last line. #Q: How do you know that for any term `T`, `Y T` is a fixed point of `T`?# A: Note that in the proof given in the previous answer, we chose `T` and then set `X = L L = (\x. T (x x)) (\x. T (x x))`. If we abstract over `T`, we get the Y combinator, `\T. (\x. T (x x)) (\x. T (x x))`. No matter what argument `T` we feed `Y`, it returns some `X` that is a fixed point of `T`, by the reasoning in the previous answer. #Q: So if every term has a fixed point, even `Y` has fixed point.# A: Right:
``````let Y = \T. (\x. T (x x)) (\x. T (x x)) in
Y Y ≡ \T. (\x. T (x x)) (\x. T (x x)) Y
~~> (\x. Y (x x)) (\x. Y (x x))
~~> Y ((\x. Y (x x)) (\x. Y (x x)))
~~> Y (Y ((\x. Y (x x)) (\x. Y (x x))))
~~> Y (Y (Y (...(Y (Y Y))...)))``````
#Q: Ouch! Stop hurting my brain.# A: Is that a question? Let's come at it from the direction of arithmetic. Recall that we claimed that even `succ`---the function that added one to any number---had a fixed point. How could there be an X such that X = X+1? That would imply that X <~~> succ X <~~> succ (succ X) <~~> succ (succ (succ (X))) <~~> succ (... (succ X)...) In other words, the fixed point of `succ` is a term that is its own successor. Let's just check that `X = succ X`:
``````let succ = \n s z. s (n s z) in
let X = (\x. succ (x x)) (\x. succ (x x)) in
succ X
≡ succ ( (\x. succ (x x)) (\x. succ (x x)) )
~~> succ (succ ( (\x. succ (x x)) (\x. succ (x x))))
≡ succ (succ X)``````
You should see the close similarity with `Y Y` here. #Q. So `Y` applied to `succ` returns a number that is not finite!# A. Yes! Let's see why it makes sense to think of `Y succ` as a Church numeral: [same definitions] succ X = (\n s z. s (n s z)) X = \s z. s (X s z) = succ (\s z. s (X s z)) ; using fixed-point reasoning = \s z. s ([succ (\s z. s (X s z))] s z) = \s z. s ([\s z. s ([succ (\s z. s (X s z))] s z)] s z) = \s z. s (s (succ (\s z. s (X s z)))) So `succ X` looks like a numeral: it takes two arguments, `s` and `z`, and returns a sequence of nested applications of `s`... You should be able to prove that `add 2 (Y succ) <~~> Y succ`, likewise for `mult`, `minus`, `pow`. What happens if we try `minus (Y succ)(Y succ)`? What would you expect infinity minus infinity to be? (Hint: choose your evaluation strategy so that you add two `s`s to the first number for every `s` that you add to the second number.) This is amazing, by the way: we're proving things about a term that represents arithmetic infinity. It's important to bear in mind the simplest term in question is not infinite: Y succ = (\f. (\x. f (x x)) (\x. f (x x))) (\n s z. s (n s z)) The way that infinity enters into the picture is that this term has no normal form: no matter how many times we perform beta reduction, there will always be an opportunity for more beta reduction. (Lather, rinse, repeat!) #Q. That reminds me, what about [[evaluation order]]?# A. For a recursive function that has a well-behaved base case, such as the factorial function, evaluation order is crucial. In the following computation, we will arrive at a normal form. Watch for the moment at which we have to make a choice about which beta reduction to perform next: one choice leads to a normal form, the other choice leads to endless reduction: let prefac = \f n. iszero n 1 (mult n (f (pred n))) in let fac = Y prefac in fac 2 = [(\f.(\x.f(xx))(\x.f(xx))) prefac] 2 = [(\x.prefac(xx))(\x.prefac(xx))] 2 = [prefac((\x.prefac(xx))(\x.prefac(xx)))] 2 = [prefac(prefac((\x.prefac(xx))(\x.prefac(xx))))] 2 = [(\f n. iszero n 1 (mult n (f (pred n)))) (prefac((\x.prefac(xx))(\x.prefac(xx))))] 2 = [\n. iszero n 1 (mult n ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred n)))] 2 = iszero 2 1 (mult 2 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred 2))) = mult 2 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] 1) ... = mult 2 (mult 1 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] 0)) = mult 2 (mult 1 (iszero 0 1 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred 0)))) = mult 2 (mult 1 1) = mult 2 1 = 2 The crucial step is the third from the last. We have our choice of either evaluating the test `iszero 0 1 ...`, which evaluates to `1`, no matter what the ... contains; or we can evaluate the `Y` pump, `(\x.prefac(xx))(\x.prefac(xx))`, to produce another copy of `prefac`. If we postpone evaluting the `iszero` test, we'll pump out copy after copy of `prefac`, and never realize that we've bottomed out in the recursion. But if we adopt a leftmost/call-by-name/normal-order evaluation strategy, we'll always start with the `iszero` predicate, and only produce a fresh copy of `prefac` if we are forced to. #Q. You claimed that the Ackerman function couldn't be implemented using our primitive recursion techniques (such as the techniques that allow us to define addition and multiplication). But you haven't shown that it is possible to define the Ackerman function using full recursion.# A. OK:
```A(m,n) =
| when m == 0 -> n + 1
| else when n == 0 -> A(m-1,1)
| else -> A(m-1, A(m,n-1))

let A = Y (\A m n. iszero m (succ n) (iszero n (A (pred m) 1) (A (pred m) (A m (pred n))))) in
```
``extract_fst ≡ \pair. pair (\x y. x)``